Main Forum Page
The Gyroscope Forum
24 March 2017 13:59
Welcome to the gyroscope forum. If you have a question about gyroscopes in general,
want to know how they work, or what they can be used for then you can leave your question here for others to answer.
You may also be able to help others by answering some of the questions on the site.
||wheels gyro effect in bike cornering
my question is how to assess the bike wheels' precession effect when in corner?
Here are my assumption in brief.
If a bike enters a curve with radius R, traveling with speed V,
that means bike is subjected to centrifugal force with magnitude m*(V*V/R),
where m is bike's mass. This force is applied at bikes center of gravity,
and in order to be compensated so the curve to be negotiated safely,
other force should balance. Leaning the bike into corner helps gravity force
provide this balance. Gravity force has magnitude m*g, where g=9.81
If the lean angle is marked with alpha, then the leaning moment
provided by gravity is m*g*cog_height*sin(alpha) should be equal to the uprigting
moment of the centrifugal force m*(V*V/R)*cog_height*cos(alpha)
This simplification gives the balance lean angle tg(alpha)=V*V/R/g,
independent both from bike's mass and center of gravity height cog_height.
I would be glad to see your opinion how wheels precession moments
could affect and correct this first approximation balance.
||3 October 2009
Answers (Ordered by Date)
||George Ivanov - 04/10/2009 21:56:33
| ||Well, here is part of mine assumptions,|
comments appriciated ...
Let's assume bike is turning left. So the wheels, forced by the frame.
Thus wheels try to precess, their angular momentum L which is pointing left,
is forced to go up, due to the ext.momentum pointing up,
this way wheel axle try to upright wheel, and bike respectively,
creating frame reaction momentum M which counteracts,
it's vector pointing backwards. This momentum can be
calculated as cause of the forced precessing wheels to left following the corner.
M= WxL , where W is angular speed via the corner : V/R,
L is (wheel inertial moment)*V/(wheel radius),
angle between them ( which sine will be used ) is 90+ alpha(lean angle)
So, may I summarize that each wheel tries to help the bike go up,
with momentum : (V/R)*((wheel inertial moment)*V/(wheel radius))*cos(alpha) ?
||menino - 19/11/2009 12:31:25
| ||the couples due to both centrifugal and gyroscopic are balanced by |
m*g*(height of c.o.g)*sin(@)
|Add an Answer >>|